I have two linear algebra texts [1, 2] with examples of the process of constructing the transition matrix that brings a matrix to its Jordan form . In each, the authors make what seems to be arbitrary selections of basis vectors via processes that do not seem algorithmic. So recently, while looking at some other calculations in linear algebra, I decided to revisit these calculations in as orderly a way as possible.
First, I needed a matrix with a prescribed Jordan form. Actually, I started with a Jordan form, and then constructed via a similarity transform on . To avoid introducing fractions, I sought transition matrices with determinant 1.
Let's begin with , obtained with Maple's JordanBlockMatrix command.
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Tools_Load Package: Linear Algebra


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The eigenvalue has algebraic multiplicity 6. There are subblocks of size 3×3, 2×2, and 1×1. Consequently, there will be three eigenvectors, supporting chains of generalized eigenvectors having total lengths 3, 2, and 1. Before delving further into structural theory, we next find a transition matrix with which to fabricate .
The following code generates random 6×6 matrices of determinant 1, and with integer entries in the interval . For each, the matrix is computed. From these candidates, one is then chosen.
After several such trials, the matrix was chosen as
for which the characteristic and minimal polynomials are
So, if we had started with just , we'd now know that the algebraic multiplicity of its one eigenvalue is 6, and there is at least one 3×3 subblock in the Jordan form. We would not know if the other subblocks were all 1×1, or a 1×1 and a 2×2, or another 3×3. Here is where some additional theory must be invoked.
The null spaces of the matrices are nested: , as depicted in Figure 1, where the vectors , are basis vectors.

Figure 1 The nesting of the null spaces



The vectors are eigenvectors, and form a basis for the eigenspace . The vectors , form a basis for the subspace , and the vectors , for a basis for the space , but the vectors are not yet the generalized eigenvectors. The vector must be replaced with a vector that lies in but is not in . Once such a vector is found, then can be replaced with the generalized eigenvector , and can be replaced with . The vectors are then said to form a chain, with being the eigenvector, and and being the generalized eigenvectors.
If we could carry out these steps, we'd be in the state depicted in Figure 2.

Figure 2 The null spaces with the longest chain determined



Next, basis vector is to be replaced with , a vector in but not in , and linearly independent of . If such a is found, then is replaced with the generalized eigenvector . The vectors and would form a second chain, with as the eigenvector, and as the generalized eigenvector.
Define the matrix by the Maple calculation
and note
The dimension of is 3, and of , 5. However, the basis vectors Maple has chosen for do not include the exact basis vectors chosen for .
We now come to the crucial step, finding , a vector in that is not in (and consequently, not in either). The examples in are simple enough that the authors can "guess" at the vector to be taken as . What we will do is take an arbitrary vector in and project it onto the 5dimensional subspace , and take the orthogonal complement as .
A general vector in is
A matrix that projects onto is
The orthogonal complement of the projection of Z onto is then . This vector can be simplified by choosing the parameters in Z appropriately. The result is taken as .
The other two members of this chain are then
A general vector in is a linear combination of the five vectors that span the null space of , namely, the vectors in the list . We obtain this vector as
A vector in that is not in is the orthogonal complement of the projection of ZZ onto the space spanned by the eigenvectors spanning and the vector . This projection matrix is
The orthogonal complement of ZZ, taken as , is then
Replace the vector with , obtained as
The columns of the transition matrix can be taken as the vectors , and the eigenvector . Hence, is the matrix
Proof that this matrix indeed sends to its Jordan form consists in the calculation
The bases for , are not unique. The columns of the matrix provide one set of basis vectors, but the columns of the transition matrix generated by Maple, shown below, provide another.
I've therefore added to my todo list the investigation into Maple's algorithm for determining an appropriate set of basis vectors that will support the Jordan form of a matrix.

References


[1] Linear Algebra and Matrix Theory, Evar Nering, John Wiley and Sons, Inc., 1963

[2] Matrix Methods: An Introduction, Richard Bronson, Academic Press, 1969




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